Frobenius Method: Solving for General Solution of 4xy'' + 2y' + y = 0 [x7/2]

[alphamu]

∑

Homework Statement

Use the method of Frobenius, constructing a power series about x = 0,
to find the general solution of this equation (retain terms up to and including those in
square brackets):

4xy'' + 2y' + y = 0 [x^7/2]

Note: the solution can be written in closed form, can you see how?

Homework Equations

The Attempt at a Solution

set y = \sum a_nx^λ+n to obtain:

4x∑(between n=0 and ∞) a_n(λ+n)(λ+n-1)x^λ+n-2 + 2∑(between n=0 and ∞) a_n(λ+n)x^λ+n-1 + ∑(between n=0 and ∞) a_nx^λ+n = 0

Now relabelling the indices using m = n-1 for the first term, m=n-1 for the second term and m=n for the third term here, in order to get a common coefficient of x^λ+m for all terms, will give:

4∑(between m=-1 and ∞) a_m+1(λ+m+1)(λ+m)x^λ+m + 2∑(between m =-1 and ∞) a_m+1(λ+m+1)x^λ+m + ∑(between m=0 and ∞) a_mx^λ+m = 0

Now look at terms when m=-1 (coefficient x^λ-1):

4[a₀(λ²-λ)] + 2[a₀λ] = 0

so 4a₀λ² - 4a₀λ +2a₀λ = 4a₀λ² - 2a₀λ = 2a₀(λ(λ-1)) = 0

with a₀ \neq 0, we have our indicial equation:

λ(λ-1) = 0 so λ₁ = 1 and λ₂ = 0

The remaining terms give:

∑(between m=0 and ∞) { a_m+1[(λ+m+1)(λ+m) + (λ+m+1)] + a_m } x^λ+m = 0

so this means ∑(between m=0 and ∞) { a_m+1[(λ+m+1)²] + a_m } x^λ+m = 0

By taking the a_m term to the other side and diving by (λ+m+1)² gives me the recurrence relation:

a_m+1 = -a_m / (λ+m+1)², m= 0, 1, 2, ...

Now looking at λ=1:

a_m+1 = -a_m / (m+2)²

and by substituting in values of m=0,1,2,3,4,5,6,7,8 will give me a₀, ... , a₈ which are the following:

a₁ = -a₀ / 4, a₂ = -a₁ / 9, a₃ = -a₂ / 16 and so on with the denominator being a squared value of (m+2) each time up until a₈ = -a₇ / 81

Here is the part I'm not sure of, I tried to use these to obtain the first part of the general solution (y₁)... by substituting into the general form y = a₀ + a₁x + a₂x² + ...

I got, y₁ = -4a₁ - 9a₂x - 16a₃x² - 25a₄x³ - 36a₅x⁴ - 49a₆x⁵ - 64a₇x⁶ - 81a₈x⁷

and then taking square root of this, i.e. (y1)^1/2, gave me the first part of the general solution in a complex form.

Is what I have done so far correct to here? I am confident it is up to the recurrence relation where I'm struggling to form a general solution after this.

I know that you need to use the other value of λ=0 to find the other part of the general solution y₂ and your final general solution will be y(x) = y1 + y2.

Again here I am not sure how to find y2 because I've used the recurrence relation to get a0, ... , a8 again but by the Frobenius method if you have two values of λ that differ by an integer (which is the case here), you examine the recurrence relation for λ₂ which I have done, but because the recurrence relation is singular at m=1 when λ=0, you seek a second solution y2 by using:

y2 = y1 ln|x| + ∑b_nx^λ+n

Would this be the correct way to find the second solution and are my working for the first solution correct?

Can anyone help me please?

 

[vela]

I got a different indicial equation: $4\lambda^2-2\lambda=0$.